Grade 8 Algebraic Identities
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1 ID : ru-8-algebraic-identities [1] Grade 8 Algebraic Identities For more such worksheets visit Answer t he quest ions (1) Solve the f ollowing using the standard identity (a+b) (a-b) = a 2 - b 2 A) B) C) D) (2) Simplif y (1ab + 4bc) 2-8ab 2 c (3) Find the value of (171.61)2 - (151.61) 2 using standard identities. (4) Solve the f ollowing using the standard identity (x + a) (x + b) = x 2 + (a + b)x + ab A) B) C) D) Choose correct answer(s) f rom given choice (5) If = 9, f ind the value of. a. 5 b. 7 c. 3 d. 6 (6) If, f ind the value of. a. 0 b. 1/2 c. 1 d. 2 (7) If 4x y 2 = 689, and xy = 20, f ind value of (2x - 5y) 2. a. 285 b. 291 c. 297 d. 289 (8) Solve the f ollowing using the standard identity a 2 - b 2 = (a+b) (a-b) a b c d. 7206
2 ID : ru-8-algebraic-identities [2] (9) If 3(a 2 + b 2 + c 2 ) = (a + b + c) 2, f ind the value of -2a + b + c. a. 0 b. -1 c. 3 d. 2 (10) If, f ind the value of x 2 - y 2. a. 3 2 b. 2 2 c. 2 d. 0 (11) If (p - 2) 2 + (q - 5) 2 + (r - 4) 2 + (s - 2) 2 + (t - 4) 2 = 0, f ind the value of pqrst. a. 303 b. 34 c. 310 d. 320 (12) If x 2 + y 2 = 8 and xy = 4, f ind the value of 3(x + y) 2-2(x - y) 2 a. 50 b. 46 c. 52 d. 48 Fill in the blanks () (46.5) 2 - (33.5) 2 = (Solve using standard identities). (14) There are two numbers such that their product is 15 and sum of the numbers is 8, the sum of their squares =. (15) There are two numbers such that sum of the numbers is 23 and their dif f erence is 9, the dif f erence of their squares = Edugain ( All Rights Reserved Many more such worksheets can be generated at
3 Answers ID : ru-8-algebraic-identities [3] (1) A) 9964 We have been asked to f ind the value of using the f ollowing identity: (a+b) (a-b) = a 2 - b 2. Let us try to think of two numbers whose sum is 106 and dif f erence is 94. Two such numbers are 100 and 6. Thus, = ( ) (100-6) = [Using the identity (a+b) (a-b) = a 2 - b 2 ] = = 9964 Theref ore, the result is B) We have been asked to f ind the value of using the f ollowing identity: (a+b) (a-b) = a 2 - b 2. Let us try to think of two numbers whose sum is 1005 and dif f erence is 995. Two such numbers are 1000 and 5. Thus, = ( ) (1000-5) = [Using the identity (a+b) (a-b) = a 2 - b 2 ] = = Theref ore, the result is
4 C) ID : ru-8-algebraic-identities [4] We have been asked to f ind the value of using the f ollowing identity: (a+b) (a-b) = a 2 - b 2. Let us try to think of two numbers whose sum is 1001 and dif f erence is 999. Two such numbers are 1000 and 1. Thus, = ( ) (1000-1) = [Using the identity (a+b) (a-b) = a 2 - b 2 ] = = Theref ore, the result is D) We have been asked to f ind the value of using the f ollowing identity: (a+b) (a-b) = a 2 - b 2. Let us try to think of two numbers whose sum is 1004 and dif f erence is 996. Two such numbers are 1000 and 4. Thus, = ( ) (1000-4) = [Using the identity (a+b) (a-b) = a 2 - b 2 ] = = Theref ore, the result is
5 (2) 1a 2 b b 2 c 2 ID : ru-8-algebraic-identities [5] We know that (a + b) 2 = a 2 + b 2 + 2ab. Now, let us start simplif ying (1ab + 4bc) 2-8ab 2 c by applying the identity (a + b) 2 = a 2 + b 2 + 2ab to the part (1ab + 4bc) 2 : (1ab + 4bc) 2-8ab 2 c = (1ab) 2 + (4bc) 2 + 2(1ab)(4bc) - 8ab 2 c = 1a 2 b b 2 c 2 + 8ab 2 c - 8ab 2 c = 1a 2 b b 2 c 2 Thus, the given expression can be simplif ied as 1a 2 b b 2 c 2. (3) 20 We have been asked to f ind the value of (171.61)2 - (151.61) 2 using standard identities. Now, (171.61) 2 - (151.61) 2 = ( )( ) [By using the identity a 2 - b 2 = (a + b)(a - b) in the numerator] 20 = = 20 Theref ore, the value of (171.61)2 - (151.61) 2 is 20.
6 (4) A) ID : ru-8-algebraic-identities [6] We have been asked to f ind the value of using the f ollowing identity: (x + a) (x + b) = x 2 + (a + b)x + ab. Let us think of two simple numbers whose sum is Two such simple numbers are 1000 and 5. Similarly, two simple numbers whose sum is 997 are 1000 and -3. Thus, = { (5)} { (-3)} = {(5) + (-3)} (5)(-3)...[Using the identity (x + a) (x + b) = x 2 + (a + b)x + ab] = (2)(1000) + (-15) = (2000) + (-15) = Theref ore, the result is B) We have been asked to f ind the value of using the f ollowing identity: (x + a) (x + b) = x 2 + (a + b)x + ab. Let us think of two simple numbers whose sum is 997. Two such simple numbers are 1000 and -3. Similarly, two simple numbers whose sum is 996 are 1000 and -4. Thus, = { (-3)} { (-4)} = {(-3) + (-4)} (-3)(-4)...[Using the identity (x + a) (x + b) = x 2 + (a + b)x + ab] = (-7)(1000) + (12) = (-7000) + (12) = Theref ore, the result is
7 C) ID : ru-8-algebraic-identities [7] We have been asked to f ind the value of using the f ollowing identity: (x + a) (x + b) = x 2 + (a + b)x + ab. Let us think of two simple numbers whose sum is 996. Two such simple numbers are 1000 and -4. Similarly, two simple numbers whose sum is 1003 are 1000 and 3. Thus, = { (-4)} { (3)} = {(-4) + (3)} (-4)(3)...[Using the identity (x + a) (x + b) = x 2 + (a + b)x + ab] = (-1)(1000) + (-12) = (-1000) + (-12) = Theref ore, the result is D) We have been asked to f ind the value of using the f ollowing identity: (x + a) (x + b) = x 2 + (a + b)x + ab. Let us think of two simple numbers whose sum is 105. Two such simple numbers are 100 and 5. Similarly, two simple numbers whose sum is 99 are 100 and -1. Thus, = { (5)} { (-1)} = {(5) + (-1)} (5)(-1)...[Using the identity (x + a) (x + b) = x 2 + (a + b)x + ab] = (4)(100) + (-5) = (400) + (-5) = Theref ore, the result is
8 (5) b. 7 ID : ru-8-algebraic-identities [8] If we assume, a = n, and b = 1/n, we can use standard algebraic identities which specif ies relation between a - b and a 2 + b 2 By using identity, (a - b) 2 = a 2 + b 2-2ab ( ) 2 = n 2 + ( 1 n ) 2-2 n 1 n ( ) 2 = 9-2 = 7 Theref ore, the value of is 7. (6) d. 2 (7) d. 289 It is given that: 4x y 2 = (1) xy = (2) Since, (a + b) 2 = a 2 + b 2 + 2ab Theref ore, (2x - 5y) 2 = {(2x) + (-5y)} 2 = (2x) 2 + (-5y) (2x) (-5y) = 4x y 2 + (-20)xy = (-20) (20)...[From equation (1) and (2)] = (-400) = 289 Thus, the value of (2x - 5y) 2 is 289.
9 (8) a ID : ru-8-algebraic-identities [9] We have been asked to f ind the value of using the f ollowing identity: a 2 - b 2 = (a + b)(a - b). Applying the identity, we can write as: ( )(86-14) = = 7200 Theref ore, the result is (9) a. 0 (10) b. 2 2 (11) d. 320 Given (p - 2) 2 + (q - 5) 2 + (r - 4) 2 + (s - 2) 2 + (t - 4) 2 = 0 It means the sum of (p - 2) 2, (q - 5) 2, (r - 4) 2, (s - 2) 2 and (t - 4) 2 is equals to 0. We know that the square of a number cannot be negative. Theref ore, the sum of these non-negative numbers (p - 2) 2, (q - 5) 2, (r - 4) 2, (s - 2) 2 and (t - 4) 2 can be zero only if all of them are also equal to zero. Now, (p - 2) 2 = 0 p - 2 = 0 p = 2 Similarly, q = 5, r = 4, s = 2, t = 4. Step 4 Thus, the value of pqrst = = 320
10 (12) d. 48 ID : ru-8-algebraic-identities [10] It is given that, x 2 + y 2 = 8 and xy = 4 Now, 3(x + y) 2-2(x - y) 2 = 3(x 2 + y 2 + 2xy) - 2(x 2 + y 2-2xy) = 3x 2 + 3y 2 + 6xy - 2x 2-2y 2 + 4xy = 1x 2 + 1y xy = 1(x 2 + y 2 ) + 10xy = 1(8) + 10(4) = 48 Thus, the value of 3(x + y) 2-2(x - y) 2 is 48. () 80 We have been asked to f ind the value of (46.5)2 - (33.5) 2 using standard identities. Now, (46.5) 2 - (33.5) 2 = b) in the numerator] 80 = = 80 ( )( ) [By using the identity a 2 - b 2 = (a + b)(a - Theref ore, the value of (46.5)2 - (33.5) 2 is 80.
11 (14) 34 ID : ru-8-algebraic-identities [11] Let s assume the two numbers be x and y. It is given that, their product is 15. Theref ore, xy = (1) Also the sum of the numbers is 8. Theref ore, x + y = 8 On squaring both sides we get: (x + y) 2 = 64 x 2 + y 2 + 2xy = 64...[Since, (x + y) 2 = x 2 + y 2 + 2xy] x 2 + y 2 + (2 15) = 64...[From eqution (1)] x 2 + y 2 = x 2 + y 2 = 34 Step 4 Thus, the sum of their squares is 34. (15) 207 Let s assume the two numbers be x and y. It is given that sum of the numbers is 23. Theref ore, x + y = (1) Also their dif f erence is 9. Theref ore, x - y = (2) Step 4 Now, the dif f erence of their squares = x 2 - y 2 = (x + y)(x - y)...[since, x 2 - y 2 = (x + y)(x - y)] = [From equation (1) and (2)] = 207 Step 5 Thus, the dif f erence of their squares is 207.
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